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Integration of an inverse sqrt composite function |
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| May16-07, 02:58 AM | #1 |
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Integration of an inverse sqrt composite function
1. The problem statement, all variables and given/known data
“Geologist A” at the bottom of a cave signals to his colleague “Geologist B” at the surface by pushing a 11.0 kg box of samples from side to side. This causes a transverse wave to propagate up the 77.0 m rope. The total mass of the rope is 14.0 kg. Take g = 9.8 m/s². How long does it take for the wave to travel from the bottom of the cave to the surface?[Hint: Find an analytic expression v(z) for the wave speed as a function of distance. Then use the fact that at any given point on the rope the time dt taken to travel a small distance dz is given by: dt=dz/v(z). Then integrate to obtain the total travel time. ] ![]() 2. Relevant equations u = mR/z T(z) = u.z.g + mB.g v(z) = (T(z)/u)^(1/2) dt=dz/v(z) z = the length of the rope = L (used for integrating) 3. The attempt at a solution v(z) = (T(z)/u)^(1/2) v(z) = ((mR/z).z.g + mB.g/(mR/z))^(1/2) v(z) = ((mR.g + mB.g)/(mR/z))^(1/2) v(z) = ([(mR.g)/(mR/z)] + [(mB.g)/(mR/z)])^(1/2) v(z) = ([(mR.g.z)/mR] + [(mB.g.z)/mR])^(1/2) v(z) = ([mR.g.z] + [(mB.g.z)/mR])^(1/2) dt=dz/v(z) dt=dz/([mR.g.z] + [(mB.g.z)/mR])^(1/2) Integration: ....f L t= | (1/([mR.g.z] + [(mB.g.z)/mR])^(1/2)).dz ....j 0 ....f L t= | 2.([mR.g.z^2/2] + [(mB.g.z^2)/mR.2])^(1/2).dz ....j 0 I think I integrated it properly but when substituted the values mB = mass of box mR = mass of rope g = 9.8 ms^2 z = 77.0 m I didn't get the correct answer of t = 2.52s |
| May16-07, 07:33 AM | #2 |
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'u' in your solution is supposed to be the mass density of the rope. It's not mR/z. The rope doesn't have a variable density, it's mR/(total length of rope), a constant. Nice problem presentation, by the way.
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| May16-07, 07:36 AM | #3 |
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How was my integration? I don't think I know how to integrate nested functions (I assume it's something like the reverse of the chain-rule?). |
| May16-07, 07:40 AM | #4 |
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Integration of an inverse sqrt composite function |
| May16-07, 07:48 AM | #5 |
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In your notation u=mR/L not mR/z.
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| May16-07, 07:49 AM | #6 |
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So:- u = mR/L = mR/77 T(z) = u.z.g + mB.g v(z) = (T(z)/u)^(1/2) dt=dz/v(z) z = the length of the rope = L (used for integrating) 3. The attempt at a solution v(z) = (T(z)/u)^(1/2) v(z) = ((u.z.g + mB.g)/u)^(1/2) v(z) = ([(u.z.g)/u] + [(mB.g)/u])^(1/2) v(z) = ([g.z] + [(mB.g)/u])^(1/2) dt=dz/v(z) dt=dz/([g.z] + [(mB.g)/u])^(1/2) Integration: ....f L t= | (1/(([g.z] + [(mB.g)/u])^(1/2)).dz ....j 0 ....f L t= | 2.([g.z] + [(mB.g)/u])^(1/2).(g.z^2)/2 ....j 0 Where: mB = mass of box mR = mass of rope g = 9.8 ms^2 z = 77.0 m Is this correct? |
| May16-07, 07:58 AM | #7 |
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Your first integral looks just fine. I don't know how you got from there to the second one. The usual way to do an integration like this is to do a change of variable. Let v=gz+mBg/u.
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| May16-07, 08:06 AM | #8 |
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So:- Integration: ....f L t= | (1/(([g.z] + [(mB.g)/u])^(1/2)).dz ....j 0 ....f L t= | 2.([(g.z^2)/2] + [(mB.g.z)/u] + C)^(1/2) ....j 0 Where: mB = mass of box mR = mass of rope g = 9.8 ms^2 z = 77.0 m How am I doing?
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| May16-07, 08:13 AM | #9 |
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Not too good. You're coming up with some pretty bizarre integration rules which aren't in the book. You have to do the change of variable thing. Eg to integrate 1/(a+bz)^(1/2) I would say v=(a+bz), so dv=b*dz. This turns the integral into 1/v^(1/2)*dv*(1/b). Now it's just integrating v^(-1/2). Does that sound familiar?
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| May16-07, 08:38 AM | #10 |
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a = g.z b = (mB.g)/u 1/(a+bz)^(1/2) v=(a+bz), so dv=b*dz. This turns the integral into 1/v^(1/2)*dv*(1/b). Integrating v^(-1/2). So:- Integration: ....f L t= | (1/(([g.z] + [(mB.g)/u])^(1/2)).dz ....j 0 ....f L t= | (1/(([g.z] + [(mB.g)/u])^(1/2)).[(mB.g)/u]dz.(1/[(mB.g)/u]) ....j 0 ....f L t= | 2.(([g.z] + [(mB.g)/u])^(1/2)).[(mB.g)/u].(1/[(mB.g)/u]) ....j 0 Where: mB = mass of box mR = mass of rope g = 9.8 ms^2 z = 77.0 m |
| May16-07, 08:56 AM | #11 |
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Take a break and clear your head. While your at it look back at integration by substitution in a calc text. I can roughly see what you are trying to do - but you still seem to be trying to do some kind of a chain rule. And the (1/b) factor in the example becomes (1/g) in the problem, right? Do you see where it's coming from? And after the integration is done and the dz is gone you should also drop the integral sign - it looks pretty confusing otherwise.
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